A roller-coaster car may be represented by a block of mass 50.0kg . The car is r

Question

A roller-coaster car may be represented by a block of mass 50.0kg . The car is released from rest at a height h = 57.0m above the ground and slides along a frictionless track. The car encounters a loop of radius R = 19.0m at ground level, as shown. As you will learn in the course of this problem, the initial height 57.0m is great enough so that the car never loses contact with the track.. What is the kinetic energy at the top of the loop? and what is the minimum initial height at which the car can be released that still allows the car to stay in contact with the track?

Explanation / Answer

form energy conservation

mgh = K.E. + mg(2R)

m x 9.8 x 57 = K.E. + m x 9.8 x (2 x 19)

K.E. = mg(h - 2R) = 9310 J

v = sqrt(2K.E./m) = sqrt(2g(h-2R))

for car to remain in contact at highest point normal force must not equal to zero.

N = mg - mv^2 /r

forlowest
we will take a case where N is just zero that will be min valye.

mg = mv^2 / R

v = sqrt(Rg)

so sqrt(Rg) = sqrt(2g(h-2R))

Rg =2 gh - 4gR

h = 2.5R = 19 x 2.5 =47.5 m

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